if the body describes a vertical circle then the minimum velocity at angle theta from lowest position is
If it is to make it around the circle then the velocity at the top must be positive. (1/2)m v^2 >/= m g (2R) if it is running around a track and can fall, then the centripetal acceleration at the top must be 1 g at least. mv^2/R >/= m g You do not say which.
To determine the minimum velocity at an angle θ from the lowest position in a vertical circle, we need to consider the concept of centripetal force.
In a vertical circular motion, the force acting on an object is the net force, which is the difference between the gravitational force and the centripetal force:
Net force = Centripetal force - Gravitational force
At the lowest position of the vertical circle, the net force acting on the object is the sum of the gravitational force and the centripetal force:
Net force = Centripetal force + Gravitational force
In order to find the minimum velocity at an angle θ, we have to ensure that the object remains in contact with the circular path at that given angle.
At the lowest position, the net force can be written as:
mv²/r = mg + mv²₀/r
Where m is the mass of the object, v is the velocity of the object at angle θ, r is the radius of the circular path, g is the acceleration due to gravity, and v₀ is the minimum velocity at the lowest point.
Simplifying the equation, we get:
mv²/r = mg + mv²₀/r
mv²/r - mv²₀/r = mg
v² - v²₀ = rg
Since we are interested in finding the minimum velocity at angle θ, the object is momentarily at rest at that point (v = 0). Thus, the equation becomes:
0 - v²₀ = rg
Rearranging the equation, we can solve for the minimum velocity at angle θ:
v₀ = sqrt(rg)
Therefore, the minimum velocity at angle θ from the lowest position in a vertical circle is given by the square root of the product of the radius of the circular path (r) and the acceleration due to gravity (g).